Suppose \( 0<|v_i|<\epsilon _0 \) , \( \epsilon _0 \)  sufficiently small, such that Theorem REF  and Theorem REF  hold. By the famous Gap-Labelling theorem, every bounded gap can be labelled by \( G_k(v)=[E_k^-,E_k^+] \)  with \( k\ne 0 \) . Without loss of generality, we only consider the bridge of \( \Sigma _i \)  at \( E_k^+ \) , denoted by \( C_k^+(v)=[E_k^+,F_k^+] \) , by the definition of bridge, \( F_k^+\in \Sigma _i \) . Thus
\(\tau (\Sigma ,E_k^+)=\frac{|F_k^+-E_k^+|}{|E_k^+-E_k^-|}.\) 
Since \( \Sigma _i \)  is a Cantor set with no isolated points, we have
\(E_k^+<F_k^+\le \sup \Sigma _i.\) 
Case 1. \( F_k^+=\sup \Sigma _i \) . Therefore
\(|N_{v,\alpha }(F_k^+)-N_{v,\alpha }(E_k^+)|=|\langle k,\alpha \rangle |_{{\begin{@align*}{1}{-1}\frac{|F_k^+-E_k^+|}{|E_k^+-E_k^-|}\ge \frac{(\frac{1}{C_0})^{2}|N_{v,\alpha }(F_k^+)-N_{v,\alpha }(E_k^+)|^2}{\epsilon _0^{\frac{1}{2}}e^{-r|2\pi k|^{{\nu }}}}\ge \frac{\gamma ^2}{C_0^2\epsilon _0^{\frac{1}{2}}}\frac{1}{e^{-r|2\pi k|^{{\nu }}}|k|^{2\tau }}.\end{@align*}}Case 2.  F_k^+<\sup \Sigma _i . Then  F_k^+  is the left boundary point of some nonempty gap  G_{k^{\prime }}(v) with  \ell (G_{k}(v))<\ell (G_{k^{\prime }}(v)) . For any nonempty gap  G_{k}(v) ,\zeta (k)=\left[\frac{1}{-r}\log [\epsilon _0^{-\frac{1}{2}}\ell (G_{k}(v))]\right]^{\frac{1}{\nu }}is well-defined. Thus by the spectral gap estimates,  \ell (G_{k}(v))\le \epsilon _0^{\frac{1}{2}}e^{-r|2\pi k|^{{\nu }}} , we have\zeta (k)\ge \zeta (k^{\prime }) \text{ and } \zeta (k)\ge \max \lbrace |k|,|k^{\prime }|\rbrace .Therefore|N_{v,\alpha }(F_k^+)-N_{v,\alpha }(E_k^+)|=|\langle k^{\prime }-k,\alpha \rangle |_{{\begin{@align*}{1}{-1}\frac{|F_k^+-E_k^+|}{|E_k^+-E_k^-|}\ge \frac{(\frac{1}{C_0})^{2}|N_{v,\alpha }(F_k^+)-N_{v,\alpha }(E_k^+)|^2}{\epsilon _0^{\frac{1}{2}}e^{-r\zeta (k)^{{\nu }}}}\ge \frac{\gamma ^2}{C_0^2\epsilon _0^{\frac{1}{2}}}\frac{1}{e^{-r\zeta (k)^{{\nu }}}|2\zeta (k)|^{2\tau }}.\end{@align*}}Notice that\tau (\Sigma )=\inf _u \tau (\Sigma ,u)=\inf _k \frac{|F_k^+-E_k^+|}{|E_k^+-E_k^-|}.Since  e^{-r\zeta (k)^{{\nu }}}|2\zeta (k)|^{2\tau }\le e^{-r|2\pi k|^{{\nu }}}|2k|^{2\tau }\rightarrow 0  when  k  large, thus there is a maximum  M  such that  e^{-r\zeta (k)^{{\nu }}}|2\zeta (k)|^{2\tau }\le M , then let  \epsilon _0\rightarrow 0 , we have\lim _{|v_i|\rightarrow 0}\tau (\Sigma _i)=+\infty .}We also need the following lemma in \cite {takase2021spectra}.\begin{lemma}[\cite {takase2021spectra}]Let  A:\mathcal {H}\rightarrow \mathcal {H}  and  B:\mathcal {H}\rightarrow \mathcal {H}  be bounded self-adjoint operators. Then|\mathrm {diam}\ \sigma (A)-\mathrm {diam}\ \sigma (B)|\le 2\mathrm {dist}_{\mathrm {Haus}}(\sigma (A),\sigma (B))\le 2\Vert A-B\Vert .\end{lemma}Now it suffices to check the conditions of Theorem \ref {astels}. For each set  \Sigma _i , by Lemma \ref {distHaus}, one has\lim \limits _{|v_i|\rightarrow 0}\mathrm {dist}_{\mathrm {Haus}}(\Sigma _i,[-2,2])=0 \text{ and } \lim \limits _{|v_i|\rightarrow 0}|\mathrm {diam}\Sigma _i-4|=0,and by Lemma \ref {thicklem},  \lim \limits _{|v_i|\rightarrow 0}\tau (\Sigma _i)=+\infty .Then there exists  \varepsilon =\varepsilon (\gamma ,\tau ,\nu ,r_0,d)>0 , such that  \sum \limits _{i=1}^b\frac{\tau (\Sigma _i)}{\tau (\Sigma _i)+1}\ge 1  and\begin{equation*}\left\lbrace \begin{aligned}&\Gamma (\Sigma _j)\le \mathrm {diam}(\Sigma _i),\ \ \forall 1\le j<i\le b,\\&\Gamma (\Sigma _i)\le \mathrm {diam}(\Sigma _1)+\cdots +\mathrm {diam}(\Sigma _{i-1}),\ \ \forall 2\le i\le b,\end{aligned}\right.\end{equation*}if  0<|v_i|<\varepsilon  for  i=1,\cdots ,b , then  \hat{\Sigma }=\Sigma _1+\Sigma _2+\cdots +\Sigma _b  is an interval.}\) 
Acknowledgments
We would like to thank Long Li, Qi Zhou and Hongyu Cheng for useful discussions, and Jiangong You for reading the earlier versions of the paper. This work is partially supported by National Key R&D Program of China (2020YFA0713300) and Nankai Zhide Foundation.

appendix

Proof of Lemma  REF
Here we prove Lemma REF  by the following quantitative Implicit Function Theorem:
Theorem 1.1 ([1]}) 
Let \( X,Y,Z \)  be Banach space, \( U\subset X \)  and \( V\subset Y \)  are neighborhoods of \( x_0 \)  and \( y_0 \)  respectively. Fix \( s,\delta >0 \)  and define \( X_s:=\lbrace x\in X ;\Vert x-x_0\Vert _X<s\rbrace  \)  and \( Y_\delta :=\lbrace y\in Y ;\Vert y-y_0\Vert _Y<\delta \rbrace  \) . Let \( \Psi \in C^{1}(U\times V,Z) \) . Suppose that \( \Psi (x_0,y_0)=0 \) , and that \( D_y\Psi (x_0,y_0)\in \mathcal {L}(Y,Z) \)  is invertible. If
\( \sup _{\overline{X_s}}\Vert \Psi (x,y_0)\Vert _Z\le \frac{\delta }{2\Vert D_y\Psi (x_0,y_0)^{-1}\Vert _{\mathcal {L}(Z,Y)}}, \) 
\( \sup _{\overline{X_s}\times \overline{Y_\delta }}\Vert \mathrm {Id}_Y-D_y\Psi (x_0,y_0)^{-1}D_y\Psi (x,y)\Vert _{\mathcal {L}(Y,Y)}\le \frac{1}{2}, \) 
then there exists \( y\in C^{1}(X_s, \overline{Y_\delta }) \) such that \( \Psi (x,y(x))=0. \)

We construct the nonlinear functional
\(\Psi :G_r^\nu (\mathbb {T},\mathrm {sl(2,{\mathbb {R}})})\times G_{r}^{\nu \,(nre)}(\eta ) \rightarrow G_{r}^{\nu \,(nre)}(\eta )\) 
by
\(\Psi (f,Y)=\mathbb {P}_{nre}[\ln \left( e^{-A^{-1}Y(\theta +\alpha )A}e^{f(\theta )}e^{Y(\theta )}\right) ],\) 
where \( \ln (\cdot ) \)  is the principal logarithm of a matrix.
One can easily check that
\( \Psi (0,0)=0,\ \Vert \Psi (f,0)\Vert \le |f|_{r}. \) 
By the definition of Fréchet derivative, We only need to consider the linear terms of \( \Psi (f,Y+Y^{\prime })-\Psi (f,Y) \) , we have
\(\Psi (f,Y+Y^{\prime }&)-\Psi (f,Y)=\mathbb {P}_{nre}\left[ \ln \left( e^{-A^{-1}(Y(\theta +\alpha )+Y^{\prime }(\theta +\alpha ))A}e^{f(\theta )}e^{Y(\theta )+Y^{\prime }(\theta +\alpha )}\right)\right.\\&\quad -\ln \left.\left( e^{-A^{-1}Y(\theta +\alpha )A}e^{f(\theta )}e^{Y(\theta )}\right)\right]\\&=\mathbb {P}_{nre}\left\lbrace \ln \left( e^{-A^{-1}(Y(\theta +\alpha )+Y^{\prime }(\theta +\alpha ))A}e^{f(\theta )}e^{Y(\theta )+Y^{\prime }(\theta )}\right)\right.\\&\quad -\ln \left( e^{-A^{-1}(Y(\theta +\alpha )+Y^{\prime }(\theta +\alpha ))A}e^{f(\theta )}e^{Y(\theta )}\right)\\&\quad +\ln \left( e^{-A^{-1}(Y(\theta +\alpha )+Y^{\prime }(\theta +\alpha ))A}e^{f(\theta )}e^{Y(\theta )}\right)\\&\quad -\ln \left.\left( e^{-A^{-1}Y(\theta +\alpha )A}e^{f(\theta )}e^{Y(\theta )}\right)\right\rbrace  .\) 

By the Baker-Campbell-Hausdorff (BCH) Formula
\( e^{X}e^Y= e^{X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}([X,[X,Y]]+[Y,[Y,X]])+\cdots },\) 
a direct calculation shows that
\(D_y\Psi (f,Y)(Y^{\prime })&=\mathbb {P}_{nre}\lbrace -A^{-1}Y^{\prime }(\theta +\alpha )A+Y^{\prime }(\theta )+E+F\rbrace ,\) 
where \(E\)  is a sum of terms of at least 2 orders in \(-A^{-1}Y(\theta +\alpha )A\) , \(f\) , \(Y^{\prime }\) , and \(F\)  is a sum of terms of at least 2 orders in \(-A^{-1}Y^{\prime }(\theta +\alpha )A\) , \(f\) , \(Y\) , but both \(E\)  and \(F\)  are only 1 order in \(Y^{\prime }\) .
Let \( f=0,Y=0 \) , thus \(E, F\)  equal zero, then
\( D_y\Psi (0,0)(Y^{\prime })=\mathbb {P}_{nre}\lbrace -A^{-1}Y^{\prime }(\theta +\alpha )A+Y^{\prime }(\theta )\rbrace . \) 
By a direct computation, which is in analogy with Lemma 1 of [2]}, we have
\( \Vert D_y\Psi (0,0)(Y^{\prime })\Vert =|-A^{-1}Y^{\prime }(\theta +\alpha )A+Y^{\prime }(\theta )|_{r}\ge \tilde{\eta }|Y^{\prime }|_{r}, \) 
where
\(\tilde{\eta }:=\frac{\eta ^3}{\eta ^2+3\Vert A\Vert ^2\eta +2\Vert A\Vert ^4}.\) 
So we have
\( \Vert D_y\Psi (0,0)^{-1}\Vert \le \tilde{\eta }^{-1}. \) 
Assume that \( \eta \ge \Vert A\Vert ^2\epsilon ^\frac{1}{9} \) , then we have
\(\tilde{\eta }&=\frac{\eta ^3}{\eta ^2+3\Vert A\Vert ^2\eta +2\Vert A\Vert ^4}\\&\ge \frac{\Vert A\Vert ^2}{3\left(\frac{\Vert A\Vert ^2}{\eta }+(\frac{\Vert A\Vert ^2}{\eta })^2+(\frac{\Vert A\Vert ^2}{\eta })^3\right)}\\&\ge \frac{1}{9}\Vert A\Vert ^2\epsilon ^{\frac{1}{3}}.\) 

Set \( s=\epsilon ,\delta =\epsilon ^{\frac{1}{2}} \) , we have
\( 2\Vert D_y\Psi (0,0)^{-1}\Vert \cdot \sup _{\overline{X_s}}\Vert \Psi (x,0)\Vert  \le 2\tilde{\eta }^{-1}\epsilon \le \delta . \) 
On the other hand,
\(\Vert D_y\Psi (0,0)(Y^{\prime })-D_y\Psi (f,Y)(Y^{\prime })\Vert &=\Vert \mathbb {P}_{nre}\lbrace E+F\rbrace \Vert \\&\le \left( 8\Vert A\Vert ^2|Y|_{r}+8\Vert A\Vert ^2|Y|_{r}\right) |Y^{\prime }|_{r}\\&\le 16\Vert A\Vert ^2\epsilon ^{\frac{1}{2}}|Y^{\prime }|_{r},\) 

so if \(\epsilon \)  sufficiently small, we have
\(\Vert \mathrm {Id}_Y-D_y\Psi (0,0)^{-1}D_y\Psi (x,y)\Vert &\le \Vert D_y\Psi (0,0)^{-1}\Vert \Vert D_y\Psi (0,0)-D_y\Psi (x,y)\Vert \\&\le \tilde{\eta }^{-1}\times 16\Vert A\Vert ^2\epsilon ^{\frac{1}{2}}\le \frac{1}{2},\) 

By Theorem REF , for \( |f|_{r}\le \epsilon \) , and \( \eta \ge \Vert A\Vert ^2\epsilon ^{\frac{1}{9}} \) , there exists \( |Y|_{r}\le \epsilon ^{\frac{1}{2}} \)  such that \( \Psi (f,Y)=0 \) , i.e.
\( \ln \left( e^{-A^{-1}Y(\theta +\alpha )A}e^{f(\theta )}e^{Y(\theta )}\right)=f^{(re)}(\theta ), \) 
which is equivalent to
\( e^{-Y(\theta +\alpha )}(Ae^{f(\theta )})e^{Y(\theta )}=Ae^{f^{(re)}(\theta )}. \) 
It is easy to check \( |f^{(re)}|_{r}\le 2\epsilon \) , which complete the proof of Lemma REF .

Proof of Lemma 
Let \(G=-\delta B^{-1}P\) , we have \(\mathrm {tr}(B^{-1}P)=0\) , hence \(G\in G^\nu _{R}(\mathbb {T}^d,\mathrm {sl}(2,{\mathbb {R}}))\) . Then we can construct \(Y\in G^\nu _{\frac{R}{2}}(\mathbb {T}^d,\mathrm {sl}(2,{\mathbb {R}}))\)  (by identifying the Fourier coefficients) such that
\(B^{-1}Y(\cdot +\alpha )B-Y=G-[G].\) 
A direct calculation shows that
\(|\widehat{Y}(n)|\le 4\gamma ^{-3}|n|^{3\tau }|\widehat{G}(n)|,\) 
thus
\(|Y|_{\frac{R}{2}}&=\sum _{n\in \mathbb {Z}^d}|\widehat{Y}(n)|e^{\frac{R}{2}(2\pi |n|)^\nu }\\&\le \sum _{n\in \mathbb {Z}^d}4\gamma ^{-3}|n|^{3\tau }e^{-\frac{R}{2}(2\pi |n|)^\nu }|\widehat{G}(n)|e^{ R(2\pi |n|)^\nu }\le D_R\delta |P|_R.\) 
Then for \(\widetilde{Z}:=e^Y\) , we have
\(\widetilde{Z}(\cdot +\alpha )^{-1}(B-\delta P(\cdot ))\widetilde{Z}(\cdot )=(B-\delta [P])+\widetilde{P}(\cdot ),\) 
where
\(\widetilde{P}(\cdot ):=\sum _{m+n\ge 2}\frac{1}{m!}(-Y(\cdot +\alpha ))^m B\frac{1}{n!}Y(\cdot )^n-\delta \sum _{m+n\ge 1}\frac{1}{m!}(-Y(\cdot +\alpha ))^m P(\cdot )\frac{1}{n!}Y(\cdot )^n. \) 
Obviously, \(|\widetilde{ Z}-\mathrm {Id}|_{\frac{R}{2}}\le 2|Y|_{\frac{R}{2}}<1\) . Since \(\sum _{m+n=k}\frac{k!}{m!n!}=2^k\) , and \(|G|_R\le \delta |P|_R\) , we get
\(|\widetilde{P}(\cdot )|_{\frac{R}{2}}\le D_R^2\delta ^2|P|_R^2+4D_R\delta ^2|P|_R^2\le (2+D_R)^2\delta ^2|P|_R^2. \) 
By a direct calculation, we can see that
\(B-\delta [P]=I+(b_0-\delta b_1)+\frac{1}{2}(b_0-\delta b_1)^2-\delta ^2 \frac{1}{2} b_1^2.\) 
Let \(P_1:=-\frac{1}{2} b_1^2+\delta ^{-2}(-\sum _{k\ge 3}\frac{1}{k!}(b_0-\delta b_1)^k+\widetilde{P})\) , note that \(b_0\)  is nilpotent, a direct computation shows (), we finish the proof of Lemma .
